Professor Erasmus has constructed a special convex pentagon $ABCDE$ that he modestly calls the "Professor-Erasmus-pentagon". The professor claims that he can cut off a smaller pentagon similar to pentagon $ABCDE$ by a straight line.

Question: Has the professor once again made one of his well-known mathematical blunders, or do such pentagons indeed exist?

Note: The five points A,B,C,D,E are all distinct. There are no tricks (like paper folding; pentagons made of rubber; etc).

  • $\begingroup$ Is it strictly convex? $\endgroup$
    – DrunkWolf
    Oct 29, 2015 at 13:36
  • $\begingroup$ @DrunkWolf: Yes. $\endgroup$
    – Gamow
    Oct 29, 2015 at 13:37

4 Answers 4


Let $A=(-30,60),B=(0,0),C=(150,0),D=(300,300),E=(225,400)$. $F=(-12,84)$ is on $AE$ and $G=(175,50)$ is on $CD$. Then $ABCDE$ is similar to $FABCG$, by a factor of $\sqrt5$.

The key point of this solution is that angles $GFA, EAB, ABC, BCD, CDE$ are all equal ($\arctan-2\approx116.57^\circ$) and $FA,AB,BC,CD$ are in geometric sequence. Then, we can rotate the pentagon so that each of those edges corresponds to the next one in the smaller pentagon.

Image of the pentagon: 1]

The same idea can be done with other values of the key angle between $90^\circ$ and $135^\circ$ and other values for the ratio of similarity.

  • $\begingroup$ +1 Providing a concrete example is much more convincing :) $\endgroup$ Oct 29, 2015 at 17:53
  • $\begingroup$ Damn, I was sure I had it. Nicely done! $\endgroup$
    – Deusovi
    Oct 29, 2015 at 17:59
  • $\begingroup$ Nice example. What's also cool I think is that you can even make all the angles the same (108°) I believe right? $\endgroup$ Oct 29, 2015 at 18:43
  • $\begingroup$ @IvoBeckers Yeah, that should be possible too. I just chose the numbers I did so that all the coordinates would be integers. $\endgroup$
    – f''
    Oct 29, 2015 at 18:59

Intuitively, I would make a polygon with sides $AB$, $BC$, $CD$, $DE$, $EA$ having lengths in increasing order $1,a,a^2,a^3,a^4$ such that the distance between, for some $a > 1$.

Then, put a point $E'$ such that $E'A = a^{-1}$ and cut the polygon on the line $E'D$. We just have to make sure that $E'D = ED = a^4$. If so, all sides will be $a^{-1},1,a,a^2,a^3$.

Edit: This is how the picture would look like (approximately, don't have any proper tool for drawing). All angles except the one between $a^3$ and $a^4$ are equal.

enter image description here

  • $\begingroup$ It is not possible like this. Firstly your first 4 sides could get point A and E a maximum of 15 units apart which is less than the 16 units side between A and E. Secondly you would also need to consider the angles in your solution to make the 2 pentagons similar. $\endgroup$ Oct 29, 2015 at 15:29
  • $\begingroup$ Yes, the angles are definitely wrong for this calculation. If $E'D = ED = 8$ but $EE' = 15.5$, then the triangle $EE'D$ has sides of length $8, 8, 15.5$. $\endgroup$ Oct 29, 2015 at 15:33
  • $\begingroup$ Yes, I realized this a bit too late. The factor 2 is a bit unfortunate... :) $\endgroup$ Oct 29, 2015 at 15:45
  • 1
    $\begingroup$ Again you didn't consider the angles. Also if you say it is possible you should be able to construct at least one such pentagon (just stating side-lengths and angles should be enough). $\endgroup$ Oct 29, 2015 at 15:50
  • $\begingroup$ This works perfectly with all equilateral triangles and $a$ satisfying $a^3 = a+1$. $\endgroup$
    – Florian F
    Jan 6 at 20:27

This is not possible.


Look at angle $C$ and find the new angle it corresponds to in the smaller pentagon. If it is one of the other angles of the larger pentagon, then every angle is congruent, so it is a regular pentagon, and this case is trivially impossible.

If $C$ and $C'$ are in the same place, then $AE=A'E'$, but the second pentagon must be smaller than the first. So $C$ must correspond to one of the newly created angles.

Of course, this applies to every angle, not just $C$. So every angle on the pentagon must either correspond to itself or a newly-created angle. Only 2 new angles can be created in any convex pentagon made from a cut through a convex pentagon. Every angle must stay in the exact same place, so the only possible solution is one with no cut at all.

Edit: This has been invalidated by f'''s example - having four congruent angles is possible in this context.

  • $\begingroup$ I also believe that is isn't possible but I don't think this deals with every corner case. "find the new angle" is not always straight forward when there are multple corners with equal angle for example $\endgroup$ Oct 29, 2015 at 16:27
  • 1
    $\begingroup$ @Ivo: If the measure of C is the same as the measure of B', then the measure of B must be the measure of A', and the measure of A must be the measure of E'... At least three of those angles have to be the same, so everything is congruent or it's flipped after the cut - I've handled both cases. $\endgroup$
    – Deusovi
    Oct 29, 2015 at 16:28

This is possible. The trick is to

Create an asymmetric Pentagon, which can be "flipped" by one cut. Use that similar allows to mirror an object.

My solution looks like this:

enter image description here

I know that

My image isn't exact, since the ratios of the sides need to be the same (e.g. ED/CB = CB/DX).

  • 11
    $\begingroup$ In your solution CD would need to get smaller to maintain the ratio which is not possible. $\endgroup$ Oct 29, 2015 at 14:36

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